All categories

3 years ago

Matemática como resolve matriz 3x5

Mathematics

2 answers:

koban [17]3 years ago

8 0

Well, 1x5=5 and 2x5=10 so 3x5=15

Black_prince [1.1K]3 years ago

7 0

So 3*5=15 hoped that helped

You might be interested in

Emily is entering a bicycle race for charity. her mother pledges $0.80 for every 0.5 mile she bikes. if emily bikes 14 miles, ho

wariber [46]

I believe the answer is $11.20

4 0

3 years ago

Read 2 more answers

Anna, Irina, Avel, and Vlad are going on a road trip. The following table shows an incomplete probability model for who will be

Tanzania [10]

Answer:3/8

Step-by-step explanation:Khan academy

4 0

3 years ago

Read 2 more answers

Your carry-on bag can weigh at most 40 pounds.Select an inequality that represents how much more weight x (in pounds) you can ad

andrey2020 [161]

Answer:

22 + x ≤ 40

x ≤ 18 pounds

Step-by-step explanation:

Maximum weight = 40 pounds

Initial weight = 22 pounds

Let Maximum weight that can be added = x

Mathematically, this means :

Initial weight + additional weight ≤ 40

That is ;

22 + x ≤ 40

Additional weight, x :

x ≤ 40 - 22

x ≤ 18 pounds

7 0

2 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

Where do parentheses go on 7+9x3-1=25

Anna35 [415]

They would go around the multiplication so it should look like 7+(9 x 3) -1=25

7 0

3 years ago